Indeed. I rushed to judgment. And my rushing is not leading to an "easy" consequence. I'm getting more interested in the question now!
–
MartyFeb 23 '12 at 21:47

2

I guess something like this works: apply complex Stirling approximation to $|\Gamma(s)| = |\Gamma(1-s)|$ to show no non-real zeros with real part outside $[0,1]$, then contour integrate to count zeros of $\Gamma(s) \pm \Gamma(1-s)$ in a rectangle $[0,1]+i[-T,T]$, and compare with the number of roots of real part $1/2$ that can again be estimated by Stirling.
–
Noam D. ElkiesFeb 23 '12 at 23:52

1

How are my approximations of zeros with 5000 digits precision explained? Checked with precision 100 in sage, pari and maple.
–
joroApr 9 '12 at 9:07

The first is easy, the second follows from Stirling's formula (this requires $n$ to be big enough, and also
requires $z$ to have imaginary part at most $4$), the third follows from the previous
two by the reflection formula for $\Gamma(z)$, the last follows by induction and by the formula
$\psi(z+1) = \psi(z) + 1/z$.
It follows that
$$\left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\Gamma'(z)}{\Gamma(z)} - \frac{d/dz (\Gamma(z) + \theta
\cdot \Gamma(1-z))}{\Gamma(z) + \theta\cdot \Gamma(1-z)} \right|$$
$$= \left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\theta \Gamma(1-z) (\psi(1-z) + \psi(z))}
{\Gamma(z) + \theta \cdot \Gamma(1-z)} \right|$$
$$ \le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)}
\oint_{C_n} \frac{1}
{|\Gamma(z) + \theta \cdot \Gamma(1-z)|}$$
$$ \le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)} \cdot \frac{1}{1/2 \Gamma(n - 1/2) + 1} \ll 1,$$
where $\theta = \pm 1$ (or anything small) and $n \ge 15$, where the final inequality holds by a huuuge margin.
It follows that
$\Gamma(z) + \theta \cdot\Gamma(1-z)$ and $\Gamma(z)$ have the same number of zeros minus the
number of poles in $C_n$. Since $\Gamma(z)$ has no zeros and poles in $C_n$, it follows that $\Gamma(z) + \theta\cdot\Gamma(1-z)$
has the same number of zeros and poles. It has exactly one pole, and thus exactly one zero.
If $\theta = \pm 1$ (and so in particular is real), by the Schwarz
reflection principle, this zero is forced to be real.
By symmetry, the same argument applies in the region $z = s + i t$ with $|t| \le 4$ and
$s \le -15$.
Combined with the above argument, this reduces the claim to $z = s + i t$ with $|s| \le 15$ and $|t| \le 4$ where the
claim can be checked directly.

Hence all the zeros outside the box $z = s + it$ with $|t| \le 4$ and $|s| \le 15$ are either in $\mathbf{R}$, or lie on the line $1/2 + i \mathbf{R}$.

EDIT To clarify, I didn't actually check that there were no ``exceptional'' zeros in the box $\pm 15 \pm 4 I$, since I presumed that the original poster had done so.
If $F(z) = \Gamma(z) - \Gamma(1-z)$, then computing the integral
$$\frac{1}{2 \pi i} \oint \frac{F'(z)}{F(z)} dz$$
around that box, one obtains (numerically, and thus exactly) $1$. There are (assuming
the OP at least computed the critical line zeros correctly) $2$ zeros in that range on the critical line. Along the real line in that range, there are $30$ poles and $25$ zeros. This means that there must be $1 + 30 - 25 = 6$ unaccounted for zeros. For such a zero
$\rho$ off the line, by symmetry one also has $\overline{\rho}$, $1 - \rho$ and
$1 - \overline{\rho}$ as zeros. Hence there must be either $1$ or $3$ pairs of zeros on the critical line, and either $1$ or $0$ quadruples of roots off the line. Varying the parameters of the integral, one can confirm there is a zero with $\rho \sim 2.7 + 0.3 i$, which is one of the four
conjugates of the root found by joro. A similar argument applies
for $\Gamma(z)+\Gamma(1-z)$. Hence:

Any zero of $\Gamma(z) - \Gamma(1-z)$ is either in $\mathbf{R}$, on the line $1/2 + i \mathbf{R}$, or is one of the four exceptional zeros $\{\rho,1-\rho,\overline{\rho},1-\overline{\rho}\}$. A similar calculation implies the same
for $\Gamma(z) + \Gamma(1-z)$, except now with an exceptional set
$\{\mu,1-\mu,\overline{\mu},1-\overline{\mu}\}$.

The first is easy, the second follows from Stirling's formula (this requires $n$ to be big enough, and also
requires $z$ to have imaginary part at most $4$), the third follows from the previous
two by the reflection formula for $\Gamma(z)$, the last follows by induction and by the formula
$\psi(z+1) = \psi(z) + 1/z$.
It follows that
$$\left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\Gamma'(z)}{\Gamma(z)} - \frac{d/dz (\Gamma(z) + \theta
\cdot \Gamma(1-z))}{\Gamma(z) + \theta\cdot \Gamma(1-z)} \right|$$
$$= \left| \frac{1}{2 \pi i} \oint_{C_n} \frac{\theta \Gamma(1-z) (\psi(1-z) + \psi(z))}
{\Gamma(z) + \theta \cdot \Gamma(1-z)} \right|$$
$$ \le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)}
\oint_{C_n} \frac{1}
{|\Gamma(z) + \theta \cdot \Gamma(1-z)|}$$
$$ \le \frac{8 |\theta| \cdot \log(n) \pi}{2 \pi \cdot \Gamma(n - 1/2)} \cdot \frac{1}{1/2 \Gamma(n - 1/2) + 1} \ll 1,$$
where $\theta = \pm 1$ (or anything small) and $n \ge 15$, where the final inequality holds by a huuuge margin.
It follows that
$\Gamma(z) + \theta \cdot\Gamma(1-z)$ and $\Gamma(z)$ have the same number of zeros minus the
number of poles in $C_n$. Since $\Gamma(z)$ has no zeros and poles in $C_n$, it follows that $\Gamma(z) + \theta\cdot\Gamma(1-z)$
has the same number of zeros and poles. It has exactly one pole, and thus exactly one zero.
If $\theta = \pm 1$ (and so in particular is real), by the Schwarz
reflection principle, this zero is forced to be real.
By symmetry, the same argument applies in the region $z = s + i t$ with $|t| \le 4$ and
$s \le -15$.
Combined with the above argument, this reduces the claim to $z = s + i t$ with $|s| \le 15$ and $|t| \le 4$ where the
claim can be checked directly.

Hence all the zeros are either in $\mathbf{R}$, or lie on the line $1/2 + i \mathbf{R}$.

EDIT To clarify, I didn't actually check that there were no ``exceptional'' zeros in the box $\pm 15 \pm 4 I$, since I presumed that the original poster had done so.
If $F(z) = \Gamma(z) - \Gamma(1-z)$, then computing the integral
$$\frac{1}{2 \pi i} \oint \frac{F'(z)}{F(z)} dz$$
around that box, one obtains (numerically, and thus exactly) $1$. There are (assuming
the OP at least computed the critical line zeros correctly) $2$ zeros in that range on the critical line. Along the real line in that range, there are $30$ poles and $25$ zeros. This means that there must be $1 + 30 - 25 = 6$ unaccounted for zeros. For such a zero
$\rho$ off the line, by symmetry one also has $\overline{\rho}$, $1 - \rho$ and
$1 - \overline{\rho}$ as zeros. Hence there must be either $1$ or $3$ pairs of zeros on the critical line, and either $1$ or $0$ quadruples of roots off the line. Varying the parameters of the integral, one can confirm there is a zero with $\rho \sim 2.7 + 0.3 i$, which is one of the four
conjugates of the root found by joro. A similar argument applies
for $\Gamma(z)+\Gamma(1-z)$. Hence:

Any zero of $\Gamma(z) - \Gamma(1-z)$ is either in $\mathbf{R}$, on the line $1/2 + i \mathbf{R}$, or is one of the four exceptional zeros $\{\rho,1-\rho,\overline{\rho},1-\overline{\rho}\}$. A similar calculation implies the same
for $\Gamma(z) + \Gamma(1-z)$, except now with an exceptional set
$\{\mu,1-\mu,\overline{\mu},1-\overline{\mu}\}$.

Wonderful. You can simplify and strengthen the proof by using a generalized Rouché's theorem. This tells us that $\Gamma(z)+\theta\cdot\Gamma(1-z)$ and $\Gamma(z)$ have the same number of zeros minus the number of poles in $C_n$ when $|\Gamma(1-z)|<|\Gamma(z)|$ holds on the boundary. This is equivalent to $\pi/|\sin(\pi z)|<|\Gamma(z)|^2$, hence it suffices to have $\pi<|\Gamma(z)|^2$ on $\partial C_n$. It seems that the last inequality holds for $n\geq 5$.
–
GH from MOFeb 25 '12 at 0:15

Very impressive, although I honestly have to say that fully understanding the proof is beyond my math skills. Still got the goosebumps from reading it though :-) The proof does induce two follow up questions: 1) could the function $\Gamma(s)^2 - \Gamma(1-s)^2$ be uniquely represented by an infinite product involving its 'complex' zeros (via Weierstrass factorization)? 2) is there a function for locating the zeros (similar to $Z(t)$ for the Riemann non trivial zeros)? Thanks.
–
AgnoFeb 25 '12 at 0:33

@Agno: Rouché's theorem is contained in basic textbooks, and this is all you need (actually a slight generalization of it). Using this you can shorten the above proof to a few lines (e.g. no integrals), see my comment above.
–
GH from MOFeb 25 '12 at 0:39

4

@GH: Rouché? Touché!
–
user631Feb 25 '12 at 3:19

@Agno: the logarithmic derivative is the tool to count zeros and it is always available.
–
Marc PalmMar 6 '12 at 18:30

I would like to expand on Guild of Pepperers's answer by noting that the zeros are essentially uniformly spaced and may easily be approximated to a high degree of accuracy. Using Stirling approximation, I obtained the formula
$$
\Gamma\left(\frac12+it\right) = \sqrt{\frac{2\pi}{1+e^{-2\pi|t|}}}\exp\left(-\frac\pi2|t|+i(t\log|t|-t+\varepsilon(t))\right),
$$
valid for real $t$, where the error $\varepsilon(t)$ is an odd, bounded, real-valued function asymptotically equal to $\frac{1}{24t}$. (Indeed, $\varepsilon(t)$ has asymptotic and convergent expansions coming from the asymptotic and convergent versions of Stirling approximation, respectively.) We then have, for $s = \frac12+it$ on the critical line,
$$
\Gamma(s)+\Gamma(1-s) = 2\sqrt{\frac{2\pi}{1+e^{-2\pi|t|}}}e^{-\frac\pi2|t|}\cos\left(t\log|t|-t+\varepsilon(t)\right),
$$
$$
\Gamma(s)-\Gamma(1-s) = 2\sqrt{\frac{2\pi}{1+e^{-2\pi|t|}}}e^{-\frac\pi2|t|}\sin\left(t\log|t|-t+\varepsilon(t)\right).
$$
One may show by means fair or foul that $t\log|t|-t+\varepsilon(t)$ is monotonically increasing for $|t|\geq1.05$, is bounded between $-0.96$ and $0.96$ for $|t|<1.05$, and is only zero when t = 0. Therefore, the zeros of $\Gamma(s)+\Gamma(1-s)$ on the critical line occur, with multiplicity one, very near those $t$ for which $t\log|t|-t$ is an odd integer multiple of $\frac{\pi}{2}$, and similarly for $\Gamma(s)-\Gamma(1-s)$ and the even integer multiples of $\frac{\pi}{2}$.

It's interesting that the number of zeros up to a given height $T$ is of the same order of magnitude, $T \log(T)$, as for the Riemann zeta function, but that these zeros have (essentially) uniform spacings rather than GUE spacings.

@Joro, you are the "Master of the Counter Example" :-) As you pointed out, the zeros of $\Gamma(s) \pm \Gamma(1-s)$ are very small, so precision of the calculation can be an issue, however with 5000 digits accuracy, your two counter examples could also easily fall in the category: (...)this reduces the claim to z=s+it with |s|≤15 and |t|≤4 where the claim can be checked directly(...). If your counter examples are correct, then my only escape is to restrict the claim to the critical strip only (similar to the $\zeta(s)$ and $\zeta^{(k)}(s)$ equivalents).
–
AgnoApr 9 '12 at 11:04

juan, how do you explain the approximations of zeros in my answer with precision 100 checked in pari, sage and maple? Have I misunderstood the question?
–
joroApr 9 '12 at 9:26

1

With mpmath I check also this zero. I think now that my parenthesis "(with some effort)" contains an error. :-/ What it is clear is that the real zeros are those of the function given. But the behavior of this function for t complex is really not simple. I will try to do an X-ray of this function. Later we will try to post it. If I know how to do it.
–
juanApr 9 '12 at 19:39

Apparently I can not post here a plot. The x-ray gives little doubt that the zeros of $\Gamma(s)-\Gamma(1-s)$ are the ones with real part $1/2$ that appeared computed in this question. The complex at $-1.69-0.30 i$ its complex conjugate the symmetrical of this with respect the critical line $2.69+0.30i$ and its complex conjugate. And then the real zeros one at $0.5$. The others real zeros can be obtained best from a real plot of the function. In the x-ray this zeros, that are very near the poles at $4$, $5$, $\dots$, can not be seen since they are contained in very short lines.
–
juanApr 10 '12 at 16:11

Of course the real zeros are symmetric with respect to 0.5 so that there are zeros near $-3$, $-4$, $\dots$
–
juanApr 10 '12 at 16:24

Given that $\Gamma(s)$ and $\Gamma(1-s)$ are complex conjugates when $\Re(s)=1/2$, it is not surprising that
$\Gamma(s)+\theta\Gamma(1-s)$ has an infinitude of zeros on the line $\Re(s)=1/2$, as long as $|\theta|=1$. The monotonicity argument given in the first answer then shows that there are no other zeros with $0<\Re(s)<1$. With the possible exception when the imaginary part of $s$ is small, the zeros for two different $\theta$ should interlace (if $\theta$ goes around the unit circle once, a zero is carried to an adjacent zero).